Specific heat, cooling and the speed of sound

Aim

To find the specific heat of a liquid by the method of mixtures. To study the cooling of a hot liquid. To find the speed of sound in air with a resonance tube.

The specific heat of a liquid by mixtures

A body of mass $m$ and specific heat $c$ that changes in temperature by $\Delta\theta$ takes in or gives out the heat $Q = mc\,\Delta\theta$. If no heat is lost to the room, the heat lost by the hot body equals the heat gained by the cold ones. That is the method of mixtures.

The apparatus. A copper block of known specific heat, a copper calorimeter with a stirrer, a thermometer, a balance and a steam heater or a beaker of boiling water. Put the calorimeter in a wooden box or felt jacket to cut the loss of heat.

The steps.

1. Find the mass $m_c$ of the empty calorimeter with its stirrer. Pour in the liquid and find the mass of the liquid $m_l$.
2. Read the temperature $\theta_0$ of the liquid.
3. Heat the copper block of mass $m_s$ in boiling water. Read its temperature $\theta_1$.
4. Move the hot block quickly into the liquid. Stir gently, and read the highest steady temperature $T$.

The calculation. The block loses heat $m_s c_s(\theta_1 - T)$. The liquid and the calorimeter gain $(m_l c_l + m_c c_c)(T - \theta_0)$. Equating these,

We want $c_l$, so $c_l$ $= \dfrac{1}{m_l}\left[\dfrac{m_s c_s(\theta_1 - T)}{T - \theta_0} - m_c c_c\right]$.

A worked example. A copper block of $150$ g at $98.0$ °C is dropped into $120$ g of oil at $25.0$ °C in a copper calorimeter of mass $60$ g. The final temperature is $38.3$ °C. Take $c_{Cu} = 385\ \text{J/(kg K)}$.

1. Heat lost by the block. $0.150\ \text{kg} \times 385\ \text{J/(kg K)}$ $\times 59.7\ \text{K}$ $= 3448$ J.
2. The temperature rise. $38.3 - 25.0 = 13.3$ K. So the oil and the calorimeter together have $\dfrac{3448\ \text{J}}{13.3\ \text{K}} = 259.2\ \text{J/K}$.
3. The calorimeter. $0.060\ \text{kg} \times 385\ \text{J/(kg K)}$ $= 23.1\ \text{J/K}$.
4. The oil. $m_l c_l = 259.2 - 23.1$ $= 236.1\ \text{J/K}$. So $c_l = \dfrac{236.1\ \text{J/K}}{0.120\ \text{kg}}$ $= 1.97 \times 10^3\ \text{J/(kg K)}$.

Cooking oil has a specific heat of about $2 \times 10^3\ \text{J/(kg K)}$, and water has $4186\ \text{J/(kg K)}$.

Precautions. Do not touch the thermometer to the calorimeter wall. Move the block quickly, so that it does not cool in the air. Do not splash. Stir gently, and take the highest reading. Start a few degrees below the room temperature and finish a few degrees above it. Then the heat gained from the room before the mixture reaches the room temperature is about equal to the heat lost to the room after.

Cooling and Newton's law

A cup of tea cools fast when it is very hot, and slowly when it is warm. Newton's law of cooling says that the rate of fall of the temperature is in proportion to the excess of the temperature $\theta$ over the surroundings $\theta_s$:

The solution is $\theta = \theta_s + (\theta_0 - \theta_s)e^{-kt}$. So the excess $\theta - \theta_s$ falls exponentially. Its logarithm is a straight line against time: $\ln(\theta - \theta_s) = \ln(\theta_0 - \theta_s) - kt$. The slope of that line is $-k$. The law holds well for a small difference of temperature, up to about $30$ K, with the surroundings still.

The steps. Heat water to about $80$ °C in a calorimeter. Place it in a draught-free place. Read the temperature every $2$ minutes. Plot $\theta$ against $t$. Then plot $\ln(\theta - \theta_s)$ against $t$.

The readings. The room is at $27.0$ °C.

The slope of the straight line is $-0.0349\ \text{min}^{-1}$, so $k = 0.0349\ \text{min}^{-1}$. The excess halves in $\dfrac{\ln 2}{k} = 19.9$ min.

Raise the room temperature. Does the curve still fall to the same level? Then double $k$. What happens to the half-time?

Comparing two liquids by cooling. Cool equal volumes of water and of a liquid in the same calorimeter. Let both cool through the same range, say from $70$ °C to $60$ °C. In the same surroundings the heat is lost at the same rate for both. So the times are in proportion to the heat capacities:

where $W$ is the heat capacity of the calorimeter. A liquid with a low specific heat cools faster.

The speed of sound with a resonance tube

A tube that is closed at one end by the surface of water, and open at the other, is a closed pipe. When a tuning fork of frequency $f$ sounds above it, the air column resonates for certain lengths. At resonance a node forms at the water surface, which is the closed end, and an antinode forms a little above the open mouth.

The antinode lies at a distance $e$ above the mouth, the end correction. For a tube of radius $r$, $e \approx 0.6r$. The two lowest resonances are at

Subtract: $L_2 - L_1 = \dfrac{\lambda}{2}$. The end correction drops out. So

We find $e$ by eliminating $\lambda$: $L_2 - 3L_1 = 2e$, so $e = \dfrac{L_2 - 3L_1}{2}$. With only $L_1$, $v = 4f(L_1 + e)$ if $e$ is known.

The steps. Clamp a glass tube vertically with a water reservoir that can be raised or lowered. Strike a fork of frequency $512$ Hz on a rubber pad. Hold it at the mouth. Move the water level slowly until the sound is loudest. Measure $L_1$ from the mouth to the water. Lower the water further to the second resonance and measure $L_2$. Repeat, and note the temperature.

The readings. Fork $512$ Hz, room at $25$ °C, a tube of radius $1.5$ cm: $L_1 = 16.0$ cm and $L_2 = 49.8$ cm.

1. The speed. $v = 2f(L_2 - L_1)$ $= 2 \times 512\ \text{Hz}$ $\times (0.498 - 0.160)\ \text{m}$ $= 346\ \text{m/s}$.
2. The end correction. $e = \dfrac{49.8 - 3 \times 16.0}{2}\ \text{cm}$ $= 0.9$ cm. The rule $0.6r = 0.9$ cm agrees.
3. The standard value. The speed of sound in air is $v \approx 331.3 + 0.606\,t\ \text{m/s}$, with $t$ in °C. At $25$ °C it is $346.4$ m/s.

Five forks. The first resonance length is $L_1 = \dfrac{v}{4f} - e$. So a graph of $L_1$ against $\dfrac{1}{f}$ is a straight line. Its slope is $\dfrac{v}{4}$ and its intercept on the $L_1$ axis is $-e$.

Raise the temperature. What happens to $L_1$ and to $L_2$? Then try a smaller fork frequency. Which resonance lengths grow?

Precautions. Strike the fork gently on a rubber pad, not on a hard surface. Keep the tube vertical and the fork above the centre of the mouth. Find the position of loudest sound with care: several tries and the mean. Read the temperature, since $v$ depends on it.

Practise with new numbers

Try it in Python

The first program fits the straight line of the first resonance length against $1/f$ for five forks. It prints the speed of sound and the end correction.

The second program fits the straight line of $\ln(\theta - \theta_s)$ against time for the cooling curve.

Common mistakes

• Forgetting the calorimeter. Its heat capacity $m_c c_c$ belongs on the side of the cold body.
• Using the starting temperature as the room temperature. The excess is measured above the room.
• Plotting $\theta$ in place of $\ln(\theta - \theta_s)$. Only the logarithm of the excess is a straight line.
• Forgetting the end correction. $L_1$ is a little less than $\lambda/4$.
• Taking $v = 4fL_1$. That ignores $e$. Use $2f(L_2 - L_1)$.

Remember these

• Mixtures: $m_s c_s(\theta_1 - T)$ $= (m_l c_l + m_c c_c)(T - \theta_0)$.
• Newton's law: $\theta - \theta_s = (\theta_0 - \theta_s)e^{-kt}$. The line of $\ln(\theta - \theta_s)$ against $t$ has the slope $-k$.
• Cooling comparison: $\dfrac{m_l c_l + W}{m_w c_w + W} = \dfrac{t_l}{t_w}$.
• Resonance tube: $v = 2f(L_2 - L_1)$, $e = \dfrac{L_2 - 3L_1}{2}$, $e \approx 0.6r$.
• $v \approx 331.3 + 0.606\,t$ m/s in air.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why is the stirrer included in the mass of the calorimeter?

2. A fork of $426$ Hz gives resonances at $19.5$ cm and $60.0$ cm. Find the speed of sound.

3. In a cooling experiment, the excess is $50$ K at time zero and $25$ K at $15$ minutes. What is $k$?

4. If the temperature of the air rises, does a resonance length of a tube become longer or shorter?

References

• NCERT, Physics Class XI, Chapter 10 (Thermal Properties of Matter), the sections on calorimetry and on Newton's law of cooling.
• NCERT, Physics Class XI, Chapter 14 (Waves), the section on resonance in a closed pipe.
• NCERT, Laboratory Manual: Physics, Class XI, the experiments on the specific heat of a liquid, the cooling curve and the speed of sound with a resonance tube.
• OpenStax University Physics Volume 1, Chapter 17 (Sound), the section on the speed of sound.